Problem 135 - A Complete Resource Book in Mathematics for JEE Main

Question

The remainder left out when \(8^{2 n}-(62)^{2 n+1}\) is divided by 9 is (A) 0 (B) 2 (C) 7 (D) 8

Step-by-Step Solution

Answer

The remainder is 2.

Step 1: Simplifying the Expression using Modulo Properties

First, we use the property in modular arithmetic that states \((a - b) \equiv (a \mod m - b \mod m) \mod m\). We need to find \((8^{2n} - (62)^{2n+1}) \mod 9\). Let's simplify each term separately modulo 9.

Step 2: Simplifying \(8^{2n} \mod 9\)

Notice that \(8 \equiv -1 \mod 9\). Therefore, \(8^{2n} \equiv (-1)^{2n} \equiv 1\) because any even power of \(-1\) results in 1. Thus, \(8^{2n} \mod 9 = 1\).

Step 3: Simplifying \(62^{2n+1} \mod 9\)

Similarly, observe that \(62 \equiv 8 \equiv -1 \mod 9\). So, \(62^{2n+1} \equiv (-1)^{2n+1} \equiv -1\). Therefore, \(62^{2n+1} \mod 9 = 8\).

Step 4: Calculating the Remainder

Now, using the results from the previous steps, substitute back into the expression: \((8^{2n} - 62^{2n+1}) \mod 9 \equiv (1 - 8) \mod 9 \equiv -7 \mod 9\). Using negative modulus properties, \(-7 \equiv 2 \mod 9\).

Step 5: Conclusion on the Remainder

Thus, the remainder when \(8^{2n} - 62^{2n+1}\) is divided by 9 is \(2\).

Key Concepts

Remainder Theorem

In modular arithmetic, the remainder theorem is a useful tool when dealing with division problems. It involves finding what remains after dividing one number by another. Think of it like slicing a cake into equal parts and seeing what's left over. When working with equations in modular arithmetic, the goal is often to find the remainder of an expression when divided by a particular number, known as the modulus.
Here's a handy principle:

• If you have an expression like \((a - b)\), you can find the remainder of this entire expression when divided by a number \(m\), using the formula: \((a \mod m - b \mod m) \mod m\).

This is particularly powerful as it allows each part of the expression to be simplified separately before combining.
This approach makes the problem-solving process much more manageable. For example, in the original exercise, we looked at the components of \(8^{2n} - 62^{2n+1}\) mod 9 to solve the problem step by step.

Powers of Integers

Working with powers of integers can sometimes be intimidating, but understanding their behavior under specific moduli can simplify things greatly. When you raise a number to a power, you're essentially multiplying it by itself several times.
But in modular arithmetic, we focus on the result of such operations modulo a number (like 9, in our exercise). Notably:

• If \(a \equiv b \mod m\), then \(a^n \equiv b^n \mod m\).

This principle tells us that if two numbers are equivalent under a modulus, raising them to the same power will not change their equivalence.
In the given problem, the clever trick was realizing that \(8 \equiv -1 \mod 9\). Therefore, \(8^{2n}\) transforms into \((-1)^{2n} \mod 9\), which, due to properties of powers of \(-1\), simplifies to 1 for any even \(n\). This made part of the problem exceptionally simpler!

Negative Modulus Properties

When dealing with negative numbers in modular arithmetic, things can seem tricky at first, but there's a straightforward way to handle them. It's important to know how to convert a negative remainder into a positive one, because in most contexts, positive remainders are preferred.
The rule is simple:

• If you find yourself with a negative number as a remainder (like \(-7\)), you can convert it by adding the modulus (like 9) to it: \(-7 + 9 = 2\).

This is because mod 9 arithmetic essentially wraps around the number line every 9 units, providing the same result whether you consider a negative or its positive counterpart.
As seen in our exercise, applying this property helped convert the result \(-7\) back into \(2\), giving us the final positive remainder. Understanding these properties helps us manage any negative numbers that crop up in modular arithmetic.